[Math] Explicit solution of a linear SDE

Question

I'd like an explicit formula as a function of $W_t$ (standard Brownian motion) and $\lambda >0$ for the solution of the following SDE:

$$\mathrm dX_t = \mathrm dW_t – \lambda X_t \,\mathrm dt$$

Someone could help me please?

Answer 1

Since the given SDE is a linear SDE we can solve this equation using a similar approach as for linear ODEs:

1. Solve the homogeneous SDE $dX_t^\circ = - \lambda \cdot X_t^\circ \, dt$: In this case that's pretty easy, we find $$X_t^\circ := X_0^\circ \cdot e^{- \lambda \cdot t}$$ as a solution of the homogeneous SDE.
2. Solve the linear SDE: Choose $X_0^\circ := 1$ and define $$Y_t := \frac{1}{X_t^\circ}= e^{\lambda \cdot t}$$ (i.e. $\frac{1}{Y_t}$ solves the homogeneous equation for inital value $Y_0=1$). Now set $Z_t := X_t \cdot Y_t$. By applying Itô's formula to $f(x,y) := x \cdot y$ we obtain $$dZ_t = Y_t \, dW_t = e^{\lambda \cdot t} \, dW_t \\ \Rightarrow Z_t = Z_0 +\int_0^t e^{\lambda \cdot s} \, dW_s$$ Thus $$X_t = \frac{Z_t}{Y_t} = e^{-\lambda \cdot t} \cdot \bigg(\underbrace{Z_0}_{X_0 \cdot Y_0=X_0}+ \int_0^t e^{\lambda \cdot s} \, dW_s \bigg)$$

---

Using this approach one can solve any linear SDE of the form

$$dX_t = (\alpha(t)+\beta(t) \cdot X_t) \, dB_t + (\gamma(t) + \delta(t) \cdot X_t) \, dt$$

where $\alpha,\beta,\gamma,\delta: [0,\infty) \to \mathbb{R}$ are determinstic coefficients.